Theorem: Reference-preserving finite regulator

At fixed $\mu ,\Lambda$, every finite observation interval $[0,T]$ and finite packet family admit a finite-dimensional prepared isometry ${\hat{V}}_t$ such that

uniformly for $t\in [0,T]$. The same bound holds for both complete complementary outputs after their common embeddings.

For $0<b<1$, it suffices to choose $m\ge 1$ and a mesh with

when $T>0$. Then $b_{\mathrm{reg}}\le b$.

Proof. In interaction picture,

On sectors containing at most $k$ particles,

An $\ell$-fold product acting on bath vacuum has norm at most $(2G{)}^{\ell }\sqrt{\ell !}$. The ordered integration simplex has volume $t^{\ell }/\ell !$. The vacuum Dyson term is consequently bounded by

The bound is uniform over source, carrier, and reference inputs.

The series converges, also after multiplication by any fixed power of $\ell$. Since each term has occupation at most $\ell$, it converges in every fixed particle-number graph norm. On these vectors the differentiated series satisfies the Schrödinger equation; uniqueness for the self-adjoint Hamiltonian identifies it with the actual evolution.

The evolution generated by $P_{m}H{P}_m$, on its projected space, has the same Dyson terms through order $m$. Inserted projections are contractions, so the difference of the two prepared isometries is at most $2Z_m(2GT)$.

Partition $[\mu ,\Lambda ]$ into intervals of width at most $\Delta$, and replace frequency multiplication by its bin midpoint value ${\tilde{h}}_B$. Then

Duhamel's formula adds at most $Tm\Delta /2$.

For an exact finite active-mode realization of this discretized model, take

It contains all coupling vectors and the complete packet space, and is invariant under ${\tilde{h}}_B$. Its orthogonal complement remains vacuum. Thus the occupation-truncated dynamics on ${\mathcal{F}}_{\mathrm{s}}(\mathfrak{f})$ is finite dimensional.

The total-occupation subspace is embedded into the product

This is an embedding, not an assertion that the total cutoff itself factorizes. The source and carrier are retained. Apply Lemma lem:isometry and partial trace.

For the explicit selection, let $x=2GT$. The first condition gives

The subsequent tail-term ratios are at most $1/2$, so $Z_m(x)\le 2^{-m}$. The second condition makes $2Z_m(x)\le b/2$; the mesh condition bounds the remaining term by $b/2$.